JEE Main 2021 · MathsMultiple choiceSingle correctMediumApplication

JEE Main 4 August 2021, Shift 1, Maths Q65

Question 65 of 90 in this shift, Maths question 5 of 30, Section A.

(1+x+x^2)^4=a_0+a_1x+a_2x^2+\ldots\ldots+a_8x^8

, then the value of

a_0+4a_2+16a_4+\ldots.+2^8a_8

is equal to

(1) $1191$
(2) $1214$
(3) $1241$ Official answer
(4) $1261$

Official answer

Option 3

NTA final key.

Chapter: Binomial Theorem
Topic: Binomial Coefficients
Idea tested: Sum of Even and Odd Coefficients

Same idea in other shifts

Asked 4× in all

3 Apr 2025, Shift 2 · Q21Let $(1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{20} x^{20}$ . If $(a_1 + a_3 + a_5 + \ldots + a_{19}) - 11a_2 = 121k$ , then…HardNumerical value
25 Jun 2022, Shift 2 · Q24If the sum of the co-efficients of all the positive even powers of $x$ in the binomial expansion of $\left(2x^3 + \frac{3}{x}\right)^{10}$ …HardNumerical value
18 Mar 2021, Shift 1 · Q66Let $(1+x+2x^2)^{20} = a_0 + a_1x + a_2x^2 + \ldots + a_{40}x^{40}$ . Then, $a_1 + a_3 + a_5 + \ldots + a_{37}$ is equal to :HardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.